How do you solve substitution problems?

For example:

a=2/5b-3
a=2b-18

or:

y=x
y= -x +2

I'm going to use a different example, cause (a) I don't like the fraction in the first one, and (b) y=x is almost too simple and I think you need to see something else.

Let's use:
2 = x + y
4 = 2x - y

The idea is to isolate a variable on one side of one of the equations. The easiest way is to find one with no coefficient on it. That is, don't use the 2x. It does not matter if it's an x or a y.

I'm going to solve for the y in the top equation. You just want to do what you'd normally do, even though the x is there: isolate to the y so that you have "y = something."

2 = x + y
I need to isolate to y, so I have to get rid of the other term, the x:
2 - x = x + y - x
2 - x = y

So we now have what y equals in the top equation. We're going to substitute this equivalent to y in the bottom equation. (Always use the other equation than the one you just used.) Remember that you have to substitute the entire (2 - x) into where y is. If it helps, put the y in parenthesis to be sure you substitute it in correctly, parenthesis and all:

4 = 2x - (y)
4 = 2x - (2 - x)

There's the substitution of the equivalent of y.
Distribute the understood negative 1:
4 = 2x - 2 + x
Combine like terms, etc. and solve for x.

Which will equal 2.

So now you have x = 2. Take that and substitute it into either equation for the x and then solve for the y.

Once you have the entire coordinate, you can put both into both equations and check that this is the correct solution. It must work in both equations. I have kept from blowing it on a test by checking my answers back in the other equation and discovered it didn't work, meaning I boo-boo'ed and was able to correct it.

(Ha! Interesting example. My y ends up being zero.)